WRITING AND GRAPHING EQUATIONS OF CONICS

GRAPHS OF RATIONAL FUNCTIONSSTANDARD FORM OF EQUATIONS OF TRANSLATED CONICS

In the following equations the point (h, k) is the vertex of the parabola and the center of the other conics.

CIRCLE (x – h) 2 + (y – k)

2 = r 2

Horizontal axis Vertical axis

PARABOLA (y – k) 2 = 4p (x – h) (x – h)

2 = 4p (y – k)

HYPERBOLA(x – h)

2 (y – k) 2

– = 1b

2a 2

(y – k) 2 (x – h)

2 – = 1

b 2a

2

ELLIPSE(x – h)

2 (y – k) 2

+ = 1a

2 b 2

(x – h) 2 (y – k)

2 + = 1

a 2b

2

Writing an Equation of a Translated Parabola

Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1).

SOLUTION

(–2, 1)(–2, 1)Choose form: Begin by sketching the parabola.Because the parabola opens to the left, it has the form

where p < 0.(y – k)

2 = 4p(x – h)

Find h and k: The vertex is at (–2, 1),so h = – 2 and k = 1.

Writing an Equation of a Translated Parabola

Write an equation of the parabola whose vertex is at (–2, 1) and whose focus is at (–3, 1).

SOLUTION

(–3, 1)(–3, 1)(–2, 1)(–2, 1)

The standard form of the equation is (y – 1) 2 = – 4(x + 2).

Find p: The distance between the vertex (–2, 1), and the focus (–3, 1) is

p = (–3 – (–2)) 2 + (1 – 1)

2 = 1

so p = 1 or p = – 1. Since p < 0, p = – 1.

Graphing the Equation of a Translated Circle

Graph (x – 3) 2 + (y + 2)

2 = 16.

SOLUTION

Compare the given equation to the standard form of the equation of a circle:

(x – h) 2 + (y – k)

2 = r 2

You can see that the graph will be a circle with center at (h, k) = (3, – 2).

(3, – 2)

The radius is r = 4

Graphing the Equation of a Translated Circle

(3 + 4, – 2 + 0) = (7, – 2)

(3 + 0, – 2 + 4) = (3, 2) (3 – 4, – 2 + 0) = (– 1, – 2)

(3 + 0, – 2 – 4) = (3, – 6)

Draw a circle through the points.

Graph (x – 3) 2 + (y + 2)

2 = 16.

SOLUTION

(– 1, – 2)

(3, – 6)

(3, 2)

(3, – 2)

r

Plot several points that are each 4 units from the center:

(7, – 2)

Writing an Equation of a Translated Ellipse

Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

SOLUTION

Plot the given points and make a rough sketch.

(x – h) 2 (y – k)

2 + = 1

a 2b

2

The ellipse has a vertical major axis,so its equation is of the form:

(3, 5)

(3, –1)

(3, 6)

(3, –2)Find the center: The center is halfwaybetween the vertices.

(3 + 3) 6 + ( –2)2

(h, k) = , = (3, 2)2

SOLUTION

(3, 5)

(3, –1)

(3, 6)

(3, –2)

Writing an Equation of a Translated Ellipse

Find a: The value of a is the distancebetween the vertex and the center.

Find c: The value of c is the distancebetween the focus and the center.

a = (3 – 3) 2 + (6 – 2)

2 = 0 + 4 2 = 4

c = (3 – 3) 2 + (5 – 2)

2 = 0 + 3 2 = 3

Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

SOLUTION

(3, 5)

(3, –1)

(3, 6)

(3, –2)

Writing an Equation of a Translated Ellipse

Find b: Substitute the values of a and cinto the equation b

2 = a 2 – c

2 .

b 2 = 4

2 – 3 2

b 2 = 7

b = 7

167+ = 1The standard form is (x – 3)

2 (y – 2) 2

Write an equation of the ellipse with foci at (3, 5) and (3, –1) and vertices at (3, 6) and (3, –2).

Graphing the Equation of a Translated Hyperbola

Graph (y + 1) 2

– = 1.(x + 1) 2

4

SOLUTION

The y 2-term is positive, so the

transverse axis is vertical. Sincea

2 = 1 and b 2 = 4, you know that

a = 1 and b = 2.

Plot the center at (h, k) = (–1, –1). Plot the vertices 1 unit above and below the center at (–1, 0) and (–1, –2).

Draw a rectangle that is centered at (–1, –1) and is 2a = 2 units high and 2b = 4 units wide.

(–1, –2)

(–1, 0)

(–1, –1)

Graphing the Equation of a Translated Hyperbola

SOLUTION

Draw the asymptotes through the corners of the rectangle.

Draw the hyperbola so that it passes through the vertices and approachesthe asymptotes.

(–1, –2)

(–1, 0)

(–1, –1)

Graph (y + 1) 2

– = 1.(x + 1) 2

4

The y 2-term is positive, so the

transverse axis is vertical. Sincea

2 = 1 and b 2 = 4, you know that

a = 1 and b = 2.

CLASSIFYING A CONIC FROM ITS EQUATION

The equation of any conic can be written in the form

Ax 2 + Bxy + Cy

2 + Dx + Ey + F = 0

which is called a general second-degree equation in x and y.

The expression B 2 – 4AC is called the discriminant

of the equation and can be used to determine which typeof conic the equation represents.

CONIC TYPES

CLASSIFYING A CONIC FROM ITS EQUATION

CONCEPT

SUMMARY

The type of conic can be determined as follows:

TYPE OF CONICDISCRIMINANT (B 2 – 4AC)

< 0, B = 0, and A = C

< 0, and either B 0, or A C

= 0

> 0

Circle

Ellipse

Parabola

Hyperbola

If B = 0, each axis is horizontal or vertical. If B 0, the axes are neither horizontal nor vertical.

Classifying a Conic

Classify the conic 2 x 2 + y

2 – 4 x – 4 = 0.

SOLUTION

Since A = 2, B = 0, and C = 1, the value of the discriminant is:

B 2 – 4 AC = 0

2 – 4 (2) (1) = – 8

Because B 2 – 4 AC < 0 and A C, the graph is an ellipse.

Help

Classifying a Conic

Classify the conic 4 x 2 – 9 y

2 + 32 x – 144 y – 5 48 = 0.

SOLUTION

Since A = 4, B = 0, and C = –9, the value of the discriminant is:

B 2 – 4 AC = 0

2 – 4 (4) (–9) = 144

Because B 2 – 4 AC > 0, the graph is a hyperbola.

Help